In a course about commutative algebra, I came across the following theorem:
$$\text{If $R$ is a noetherian ring, then $R[X_1,\dots,X_n]$ is noetherian.}$$
I can't help but wonder, if the converse is also true. Namely, if $R[X_1,\dots,X_n]$ is noetherian, then $R$ is noetherian itself?
Intuitively this would make sense for me, but I know that modules sometimes behave in some strange ways. It seems fishy, since I can't find this statement anywhere online or in my notes. And if such a statement were true, I should be able to find it somewhere...
An intuitive reason for this to be true, would be that if $R$ is non-noetherian, one can find a sequence of ideals $I_1 \subset I_2 \subset \dots \subset R$, which never come to a halt. Now to generate a corresponding sequence in the $R[X]$, we just take the ideals $I_k'$, generated by $X$ and the generators of $I_k$, which shows that $R[X]$ cannot be noetherian.
Another idea would be via morphism $f: R[X_1,\dots,X_n] \rightarrow R$ ; $ f \mapsto f(0,\dots,0)$. This is clearly surjective, therefore induces an isomorphism $R[X_1,\dots,X_n]/\ker(f) \simeq R$, which in turn implies that $R$ is noetherian.
Does this indeed hold, and do the "proofs" work?